C, and D. They’re parallel so they’re never going to cross.
expenses is greater in Jan and Feb
Jan −85.6 Loss
Feb −162.44 Loss
Mar 0
June 306.77 Gain
July 301.5 Gain
Aug 337.88 Gain
Angle C equals 63
Add angle 17 plus 100 and the subtract the answer from 180. Because a triangles are always 180 degrees
Answer:
•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25
•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25
cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗
sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5
Answer:
13.98 in²
Step-by-step explanation:
I don't understand it, either.
Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which <u>is not</u> an answer choice.
__
The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which <u>is</u> an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.
The area of a segment is given by the formula ...
A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.
Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...
A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²
Rounded to hundredths, this is ...
≈ 13.98 in²
